There are a few ways to approach this, but here's one way: restrict $V$ to a finite-dimensional space.
Take bases $(T_1 v_1, \ldots, T_1 v_n)$ of $\operatorname{range} T_1$ and $(T_2 w_1, \ldots, T_2 w_m)$ of $\operatorname{range} T_2$. Then let
$$V' = \operatorname{span} \lbrace v_1, \ldots, v_n, w_1, \ldots, w_m \rbrace \le V.$$
Note that $V'$ is finite-dimensional, and
\begin{align*}
\operatorname{range} (T_1|_{V'}) &= \operatorname{span} \lbrace T_1 v_1, \ldots, T_1 v_n, T_1 w_1, \ldots, T_1w_m \rbrace \\
&\supseteq \operatorname{span} \lbrace T_1 v_1, \ldots, T_1 v_n \rbrace \\
&= \operatorname{range}(T_1) \\
&\supseteq \operatorname{range}(T_1|_{V'}).
\end{align*}
Hence $\operatorname{range}(T_1) = \operatorname{range}(T_1|_{V'})$. Similarly, $\operatorname{range}(T_2) = \operatorname{range}(T_2|_{V'})$.
As for the nullspaces,
$$\operatorname{null} T_1|_{V'} = \operatorname{null} T_1 \cap V' \subseteq \operatorname{null} T_2 \cap V' \subseteq \operatorname{null} T_2|_{V'}.$$
Since you can prove the result on finite-dimensional spaces, you now know that
$$\operatorname{dim} \operatorname{range} T_1|_{V'} \ge \operatorname{dim} \operatorname{range} T_2|_{V'}.$$
But, since restriction to $V'$ didn't affect the range,
$$\operatorname{dim} \operatorname{range} T_1 \ge \operatorname{dim} \operatorname{range} T_2.$$
\ge
for $\ge$ and\le
for $\le$. Also, to make your own operators (like $\operatorname{dim}$ and $\operatorname{null}$), use the command\operatorname{null}
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