[proofplan] We prove finite-dimensionality of $U$ by constructing a basis iteratively: start from the empty set and keep adding vectors from $U$ that are not in the current span. The [Steinitz Exchange Lemma](/theorems/373) bounds the size of any linearly independent subset of $V$ by $\dim V = n$, so the process terminates, yielding a basis for $U$ with at most $n$ elements. For proper subspaces, if $\dim U = n$ then any basis for $U$ would be a basis for $V$ by [Properties of Finite Dimensional Bases](/theorems/374), contradicting $U \neq V$. [/proofplan] [step:Build a basis for $U$ by iterative extension] Construct a linearly independent subset of $U$ as follows. Start with the empty set. If the current independent set $\{u_1, \ldots, u_k\}$ does not span $U$, choose $u_{k+1} \in U \setminus \langle u_1, \ldots, u_k \rangle$. The augmented set $\{u_1, \ldots, u_{k+1}\}$ is linearly independent: if $\sum_{i=1}^{k+1}\lambda_i u_i = \mathbf{0}$ with $\lambda_{k+1} \neq 0$, then $u_{k+1} = -\lambda_{k+1}^{-1}\sum_{i=1}^{k}\lambda_i u_i \in \langle u_1, \ldots, u_k \rangle$, contradicting the choice of $u_{k+1}$; hence $\lambda_{k+1} = 0$, and the independence of $\{u_1, \ldots, u_k\}$ forces all $\lambda_i = 0$. By the [Steinitz Exchange Lemma](/theorems/373), any linearly independent subset of $V$ has at most $n = \dim V$ elements. Hence the process terminates after at most $n$ steps, producing a basis $\{u_1, \ldots, u_d\}$ for $U$ with $d \leq n$. In particular, $U$ is finite-dimensional with $\dim U \leq n$. [/step] [step:Prove strict inequality for proper subspaces] Suppose $U \subsetneq V$ and assume for contradiction that $\dim U = n = \dim V$. Let $\{u_1, \ldots, u_n\}$ be a basis for $U$. This is a linearly independent subset of $V$ with $n$ elements. By [Properties of Finite Dimensional Bases](/theorems/374) part (ii), any linearly independent subset of $V$ with $n = \dim V$ elements is a basis for $V$. Hence $\langle u_1, \ldots, u_n \rangle = V$. But $\langle u_1, \ldots, u_n \rangle = U \neq V$, a contradiction. Therefore $\dim U < n$. [/step]