[proofplan] Exactness identifies $U$ with $\operatorname{im}(f)=\ker(g)$ and identifies $W$ with the quotient information carried by $g$. The main argument is a basis-lifting calculation: a basis of $\ker(g)$, together with chosen lifts of a basis of $W$, forms a basis of $V$. This proves the dimension formula directly when $U$ and $W$ are finite-dimensional, and the remaining cases follow by applying the same calculation to bases extracted from the finite-dimensional ambient space $V$. [/proofplan]

[step:Translate exactness into the two structural facts used below] Since the sequence is exact, the map $f:U \to V$ is injective, the map $g:V \to W$ is surjective, and \begin{align*} \operatorname{im}(f)=\ker(g). \end{align*} Because $f$ is injective, the map \begin{align*} f:U &\to \operatorname{im}(f) \end{align*} is an isomorphism of $k$-vector spaces. Hence $U$ is finite-dimensional if and only if $\operatorname{im}(f)$ is finite-dimensional, and in that case \begin{align*} \dim_k U=\dim_k \operatorname{im}(f)=\dim_k \ker(g). \end{align*} [/step]

[step:Prove the basis-lifting dimension calculation]Let $K:=\ker(g)$, viewed as a $k$-vector subspace of $V$. Suppose first that $K$ and $W$ are finite-dimensional. Choose a basis \begin{align*} (a_1,\dots,a_r) \end{align*} of $K$, where $r=\dim_k K$. Choose a basis \begin{align*} (c_1,\dots,c_s) \end{align*} of $W$, where $s=\dim_k W$. Since $g:V \to W$ is surjective, for each index $j \in \{1,\dots,s\}$ choose an element $b_j \in V$ such that \begin{align*} g(b_j)=c_j. \end{align*} We claim that \begin{align*} (a_1,\dots,a_r,b_1,\dots,b_s) \end{align*} is a basis of $V$. To prove spanning, let $v \in V$. Since $(c_1,\dots,c_s)$ spans $W$, there are scalars $\beta_1,\dots,\beta_s \in k$ such that \begin{align*} g(v)=\sum_{j=1}^{s}\beta_j c_j. \end{align*} Define \begin{align*} v_0:=v-\sum_{j=1}^{s}\beta_j b_j \in V. \end{align*} By linearity of $g$ and the choice of the $b_j$, \begin{align*} g(v_0)=g(v)-\sum_{j=1}^{s}\beta_j g(b_j)=g(v)-\sum_{j=1}^{s}\beta_j c_j=0. \end{align*} Thus $v_0 \in K$. Since $(a_1,\dots,a_r)$ spans $K$, there are scalars $\alpha_1,\dots,\alpha_r \in k$ such that \begin{align*} v_0=\sum_{i=1}^{r}\alpha_i a_i. \end{align*} Therefore \begin{align*} v=\sum_{i=1}^{r}\alpha_i a_i+\sum_{j=1}^{s}\beta_j b_j, \end{align*} so the displayed list spans $V$. To prove linear independence, suppose scalars $\alpha_1,\dots,\alpha_r,\beta_1,\dots,\beta_s \in k$ satisfy \begin{align*} \sum_{i=1}^{r}\alpha_i a_i+\sum_{j=1}^{s}\beta_j b_j=0. \end{align*} Applying the linear map $g$ gives \begin{align*} 0=\sum_{i=1}^{r}\alpha_i g(a_i)+\sum_{j=1}^{s}\beta_j g(b_j). \end{align*} Each $a_i$ lies in $K=\ker(g)$, so $g(a_i)=0$, while $g(b_j)=c_j$. Hence \begin{align*} \sum_{j=1}^{s}\beta_j c_j=0. \end{align*} Since $(c_1,\dots,c_s)$ is linearly independent, every $\beta_j$ is $0$. The original relation becomes \begin{align*} \sum_{i=1}^{r}\alpha_i a_i=0. \end{align*} Since $(a_1,\dots,a_r)$ is linearly independent, every $\alpha_i$ is $0$. Thus the displayed list is linearly independent, and therefore it is a basis of $V$. Consequently $V$ is finite-dimensional and \begin{align*} \dim_k V=r+s=\dim_k K+\dim_k W. \end{align*}[/step]

[guided]The point of this step is to show exactly how a short exact sequence decomposes a basis of the middle term into two parts: one part lying in the kernel of $g$, and one part mapping onto a basis of $W$. Let $K:=\ker(g)$. Assume $K$ and $W$ are finite-dimensional. Choose a basis \begin{align*} (a_1,\dots,a_r) \end{align*} of $K$, where $r=\dim_k K$, and choose a basis \begin{align*} (c_1,\dots,c_s) \end{align*} of $W$, where $s=\dim_k W$. Since $g:V \to W$ is surjective, every basis vector $c_j$ has at least one preimage in $V$. For each $j \in \{1,\dots,s\}$, choose $b_j \in V$ satisfying \begin{align*} g(b_j)=c_j. \end{align*} We prove that \begin{align*} (a_1,\dots,a_r,b_1,\dots,b_s) \end{align*} is a basis of $V$. First, let $v \in V$. Since the $c_j$ span $W$, there are scalars $\beta_1,\dots,\beta_s \in k$ such that \begin{align*} g(v)=\sum_{j=1}^{s}\beta_j c_j. \end{align*} Subtract from $v$ the corresponding lifted combination: \begin{align*} v_0:=v-\sum_{j=1}^{s}\beta_j b_j. \end{align*} This subtraction is designed to kill the image under $g$. Indeed, by linearity of $g$, \begin{align*} g(v_0)=g(v)-\sum_{j=1}^{s}\beta_j g(b_j). \end{align*} Using $g(b_j)=c_j$, this becomes \begin{align*} g(v_0)=g(v)-\sum_{j=1}^{s}\beta_j c_j=0. \end{align*} Hence $v_0 \in \ker(g)=K$. Since the $a_i$ span $K$, there are scalars $\alpha_1,\dots,\alpha_r \in k$ with \begin{align*} v_0=\sum_{i=1}^{r}\alpha_i a_i. \end{align*} Substituting the definition of $v_0$ gives \begin{align*} v=\sum_{i=1}^{r}\alpha_i a_i+\sum_{j=1}^{s}\beta_j b_j. \end{align*} Thus the chosen vectors span $V$. Now suppose there is a linear relation \begin{align*} \sum_{i=1}^{r}\alpha_i a_i+\sum_{j=1}^{s}\beta_j b_j=0. \end{align*} Apply $g$ to the relation. Since each $a_i$ lies in $K=\ker(g)$, we have $g(a_i)=0$. Since $g(b_j)=c_j$, the relation becomes \begin{align*} \sum_{j=1}^{s}\beta_j c_j=0. \end{align*} The vectors $c_1,\dots,c_s$ are linearly independent, so $\beta_j=0$ for every $j$. The original relation is then \begin{align*} \sum_{i=1}^{r}\alpha_i a_i=0. \end{align*} The vectors $a_1,\dots,a_r$ are linearly independent, so $\alpha_i=0$ for every $i$. Therefore all coefficients in the original relation vanish, proving linear independence. We have shown that the list has $r+s$ vectors and is a basis of $V$. Hence $V$ is finite-dimensional and \begin{align*} \dim_k V=r+s=\dim_k K+\dim_k W. \end{align*}[/guided]

[step:Handle the case where $U$ and $W$ are finite-dimensional] Assume $U$ and $W$ are finite-dimensional. By the first step, $K=\ker(g)=\operatorname{im}(f)$ is finite-dimensional and \begin{align*} \dim_k K=\dim_k U. \end{align*} Applying the basis-lifting calculation to $K$ and $W$ shows that $V$ is finite-dimensional and \begin{align*} \dim_k V=\dim_k K+\dim_k W=\dim_k U+\dim_k W. \end{align*} [/step]

[step:Handle the cases where $V$ is finite-dimensional] Assume $V$ is finite-dimensional. Every subspace of a finite-dimensional vector space is finite-dimensional, so $K=\ker(g)$ is finite-dimensional. Choose a basis $(a_1,\dots,a_r)$ of $K$ and extend it to a basis \begin{align*} (a_1,\dots,a_r,b_1,\dots,b_s) \end{align*} of $V$. We now prove that $(g(b_1),\dots,g(b_s))$ is a basis of $W$. First, let $w \in W$. Since $g:V \to W$ is surjective, there exists $v \in V$ such that $g(v)=w$. Because $(a_1,\dots,a_r,b_1,\dots,b_s)$ is a basis of $V$, there are scalars $\alpha_1,\dots,\alpha_r,\beta_1,\dots,\beta_s \in k$ such that \begin{align*} v=\sum_{i=1}^{r}\alpha_i a_i+\sum_{j=1}^{s}\beta_j b_j. \end{align*} Applying $g$ and using $g(a_i)=0$ for every $i$ gives \begin{align*} w=g(v)=\sum_{j=1}^{s}\beta_j g(b_j). \end{align*} Thus the vectors $g(b_1),\dots,g(b_s)$ span $W$. To prove linear independence, suppose scalars $\beta_1,\dots,\beta_s \in k$ satisfy \begin{align*} \sum_{j=1}^{s}\beta_j g(b_j)=0. \end{align*} Then \begin{align*} g\Bigl(\sum_{j=1}^{s}\beta_j b_j\Bigr)=0, \end{align*} so $\sum_{j=1}^{s}\beta_j b_j \in K$. Since $(a_1,\dots,a_r,b_1,\dots,b_s)$ is a basis of $V$ and $(a_1,\dots,a_r)$ is a basis of $K$, the vector $\sum_{j=1}^{s}\beta_j b_j$ has zero coordinates along the $b_j$-directions only when every $\beta_j$ is $0$. Therefore $g(b_1),\dots,g(b_s)$ are linearly independent. Hence they form a basis of $W$, so $W$ is finite-dimensional and \begin{align*} \dim_k W=s. \end{align*} Also $\dim_k K=r$, and since $f:U \to K$ is an isomorphism, $\dim_k U=\dim_k K=r$. Therefore, whenever $V$ is finite-dimensional and either $U$ or $W$ is already known to be finite-dimensional, the remaining space is finite-dimensional and \begin{align*} \dim_k V=r+s=\dim_k U+\dim_k W. \end{align*} In particular, this covers both cases where $U$ and $V$ are finite-dimensional and where $V$ and $W$ are finite-dimensional. [/step]

[step:Conclude the dimension identity in all allowed cases] The hypothesis says that at least two of $U$, $V$, and $W$ are finite-dimensional. If the finite-dimensional spaces are $U$ and $W$, the third step proves that $V$ is finite-dimensional and gives the desired identity. If $V$ is one of the finite-dimensional spaces, the preceding step proves that the remaining space is finite-dimensional and gives the same identity. Hence in every case covered by the hypothesis, \begin{align*} \dim_k V=\dim_k U+\dim_k W. \end{align*} This proves the theorem. [/step]